Understanding the calculation of expectation value

Question

The expectation value (in sense of discrete probability) can be thought of as $$ \left<a\right>=\frac{1}{N}\sum\limits^{N}{Â }\psi $$ where $N$ is the number of experiments. As the number of experiments go to infinity the expression converts into an integral. $$ \begin{array}{l} \left<a\right> =\int_{-\infty }^{\infty }{{\psi }^{*}}Â\psi dx\\ \end{array} $$ The lower and upper bound are for $\psi$. Is the existence of $ {\psi }^{*} $ simply due to the integral of a complex function or is there more to it?

Answer 1

This is not correct in more than one way.

First, increasing the number of trials does not transform the sum into an integral. If you throw a die $10^{20}$ time the average value of the throw is still $$ \frac{1}{10^{20}}\sum_n n N(n) $$ where $N(n)$ is the number of times you observe outcome $n$ on the die, i.e it still a sum over $6$ possible outcomes. More generally, $$ \langle a\rangle =\frac{1}{N}\sum_a a A(a). \tag{1} $$ where $A(a)$ is the number of times you get the outcome $a$, and the sum is over all possible outcomes. The sum remains discrete even as you increase the number of experiments provided that the possible outcomes remain discrete, as in throwing a single die.

The key difference between something like Eq.(1) and $$ \int dx P(x) x \tag{2} $$ is the nature of the outcomes, either discrete as in (1) or continuous like $x$ in (2).

In (2), the function $P(x)$ is a probably density, i.e.a non-negative function of $x$. Note that it is a density in the sense that, if $\int dx P(x)=1$, where the integration is over the entire range of $x$, the $P(x_0)dx$ is the probability of getting the $x$ values to lie between $x_0\pm dx/2$, i.e in the range $x_0-dx/2$ to $x_0+dx/2$. We use (2) to compute - say - the average position because the possible positions are a continuous set of outcomes.

The role of this non-negative probability density $P(x)$ is in quantum mechanics played by $\psi^*(x)\psi(x)$, which is also non-negative. For bound states we further require that $\psi(x)^*\psi(x) <\infty$ so we can rescale $\psi(x)$ so $\int dx \psi^*(x)\psi(x)=1$.

The expression you have involving $\hat A\psi$ is NOT an average value. In fact it’s just the state $\psi$ which results from the application of $\hat A$.

Answer 2

Is the existence of ψ∗ simply due to the integral of a complex function or is there more to it?

I shall use Dirac notation:

Given that the state of some quantum system is described by the ket, $\lvert \psi \rangle$, the wave function represented in position space is defined as $\psi(x) = \langle x \lvert \psi \rangle$, but I shall continue with just the abstract ket. We are guaranteed the existence of the complex transpose of $\lvert \psi \rangle$ by the fact that the Hilbert space that $\lvert \psi \rangle$ lives in is complex. More formally, the complex transpose of our state, $\lvert \psi \rangle$* = $\langle \psi \rvert$ is also called the dual of $\lvert \psi \rangle$ and is guaranteed to exist. This might also be a helpful conversation.